HDU6322 Problem D. Euler Function【欧拉函数+数学规律】

Problem D. Euler Function
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1960 Accepted Submission(s): 1567

Problem Description
In number theory, Euler’s totient function φ(n) counts the positive integers up to a given integer n that are relatively prime to n. It can be defined more formally as the number of integers k in the range 1≤k≤n for which the greatest common divisor gcd(n,k) is equal to 1.
For example, φ(9)=6 because 1,2,4,5,7 and 8 are coprime with 9. As another example, φ(1)=1 since for n=1 the only integer in the range from 1 to n is 1 itself, and gcd(1,1)=1.
A composite number is a positive integer that can be formed by multiplying together two smaller positive integers. Equivalently, it is a positive integer that has at least one divisor other than 1 and itself. So obviously 1 and all prime numbers are not composite number.
In this problem, given integer k, your task is to find the k-th smallest positive integer n, that φ(n) is a composite number.

Input
The first line of the input contains an integer T(1≤T≤100000), denoting the number of test cases.
In each test case, there is only one integer k(1≤k≤109).

Output
For each test case, print a single line containing an integer, denoting the answer.

Sample Input
2
1
2

Sample Output
5
7

Source
2018 Multi-University Training Contest 3

问题链接：HDU6322 Problem D. Euler Function
问题简述：（略）
问题分析：欧拉函数问题，可以通过打表找规律。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* HDU6322 Problem D. Euler Function */#include <bits/stdc++.h>int main()
{int t, n;scanf("%d", &t);while(t-- && ~scanf("%d", &n))printf("%d\n", n == 1 ? 5 : n + 5);return 0;
}

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HDU6322 Problem D. Euler Function【欧拉函数+数学规律】

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